Application des propriétés algébriques pour le calcul de modules - Corrigé

Énoncé

Déterminer le module des complexes suivants.

1.  `z_1=(5+2i)(10-3i)`     

2.  `z_2=(1-4i)^5`   

3.  `z_3=\frac{2i-\sqrt{5}}{3-4i}`

Solution

1.  On a : \(\begin{align*} \left\vert z_1 \right\vert = \left\vert (5+2i)(10-3i) \right\vert = \left\vert 5+2i \right\vert \times \left\vert 10-3i \right\vert \end{align*}\)
or \(\left\vert 5+2i \right\vert =\sqrt{5^2+2^2} =\sqrt{25+4} =\sqrt{29}\)
et \(\left\vert 10-3i \right\vert =\sqrt{10^2+(-3)^2} =\sqrt{100+9} =\sqrt{109}\)
donc \(\left\vert z_1 \right\vert =\sqrt{29} \times \sqrt{109} = \sqrt{3\,161}\) .  

2.  On a : \(\begin{align*} \left\vert z_2 \right\vert = \left\vert (1-4i)^5 \right\vert = \left\vert 1-4i \right\vert^5 \end{align*}\)
or \(\left\vert 1-4i \right\vert =\sqrt{1^2+(-4)^2} =\sqrt{1+16} =\sqrt{17}\)
donc  \(\left\vert z_2 \right\vert = \left(\sqrt{17} \right)^5\) .

3.  On a : \(\begin{align*} \left\vert z_3 \right\vert =\left\vert \frac{2i-\sqrt{5}}{3-4i} \right\vert =\frac{\left\vert 2i-\sqrt{5} \right\vert}{\left\vert 3-4i \right\vert} \end{align*}\)
or \(\left\vert 2i-\sqrt{5} \right\vert =\sqrt{(-\sqrt{5})^2+2^2} =\sqrt{5+4} =\sqrt{9} =3\)
et \(\left\vert 3-4i \right\vert =\sqrt{3^2+(-4)^2} =\sqrt{9+16} =\sqrt{25} =5\)
donc  \(\left\vert z_3 \right\vert =\dfrac{3}{5}\) .